package com.xsherl.leetcode.solution;

public class RegularExpressionMatching {

    /**
     * 动态规划
     * f[i][j] = {
     *     f[i - 1][j - 1], p[j - 1] != '*' && s[i] == p[j]
     *     f[i - 1][j - 1], p[j - 1] == '*' && s[i] != p[j]
     * }
     */
    public boolean isMatch(String s, String p) {
        int m = s.length();
        int n = p.length();
        boolean[][] f = new boolean[m + 1][n + 1];
        f[0][0] = true;
        for (int i = 0; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (p.charAt(j - 1) == '*') {
                    f[i][j] = f[i][j - 2];
                    if (matches(s, p, i, j - 1)) {
                        f[i][j] = f[i][j] || f[i - 1][j];
                    }
                } else {
                    if (matches(s, p, i, j)) {
                        f[i][j] = f[i - 1][j - 1];
                    }
                }
            }
        }
        return f[m][n];
    }

    public boolean matches(String s, String p, int i, int j){
        if (i == 0){
            return false;
        }
        return p.charAt(j - 1) == '.' || s.charAt(i - 1) == p.charAt(j - 1);
    }

    public static void main(String[] args) {
        String s = "aa", p = "a";
        boolean match = new RegularExpressionMatching().isMatch(s, p);
        System.out.println(match);
    }

}
